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I passed by this problem in Abbott's "Understanding Analysis", page 11:

Produce an infinite collection of sets $A_1, A_2, A_3,...$ with the property that every $A_i$ has an infinite number of elements, $A_i ∩ A_j = \phi$ for all $i \not= j$ , and $\bigcup_{i=1}^{\infty} A_i = \mathbf N$

  1. Finding the solution was not difficult conceptually (that is, if my attempt is correct), but I have no idea how to formulate it. My idea is as follows:

    If the question asked for two sets, an answer would be the sets of even and odd numbers. If they're $3$ sets, I would then give $A_1$ all the numbers starting from $1$ and increment $3$ to get $A_1 = \{1,4,7,...\}$, similarly, for $A_2$, starting from $2$ and incrementing by $3$, etc... For $n$ sets, $A_i = \{i, i+n,i+2n,...\}$.

    However, for an infinite number of sets, formulating a more formal solution is confusing to me.

  2. My other question is this: Is this the only way to solve this problem, and why?

Omar
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2 Answers2

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Hint: use the Fundamental Theorem of Artithmetic and Euclid's proof that there are infinitely many primes. Good luck!

Integrand
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  • I failed to get this hint anywhere. Could you elaborate? – Omar Jan 20 '21 at 20:32
  • For instance: let $A_1= {1}\cup 2\mathbb{N}$. Then let $A_2 = 3\mathbb{N}\setminus A_1$, $A_3 = 5 \mathbb{N}\setminus (A_1\cup A_2)$, etc. – Integrand Jan 20 '21 at 21:20
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There are many ways to do this. My favorite is to give $A_1$ every second natural number, $$ A_1=\{1,3,5,7,\dots\}, $$ then give $A_2$ every second of the remaining numbers, $$ A_2=\{2,6,10,14,\dots\}, $$ then give $A_3$ every second of the (still) remaining numbers, $$ A_3=\{4,12,20,28,\dots\} $$ and so on. Equivalently, $$ A_n=\{2^{n-1}k:k\text{ is odd}\}. $$ Another equivalent formulation is that $A_n$ consists of those positive integers whose binary expression ends with exactly $n-1$ zeros.

Andreas Blass
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